Probability that exactly one of the event E or F occurs is

\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}\)

Now,

\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}-{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}\)

Since \(\displaystyle{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}={0}\)

\(\displaystyle\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}\)

Also, \(\displaystyle{P}{\left({E}{F}^{{c}}\right)}={P}{\left({E}\right)}-{P}{\left({E}{F}\right)}\) and \(\displaystyle{P}{\left({E}^{{c}}{F}\right)}={P}{\left({F}\right)}-{P}{\left({E}{F}\right)}\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}\right)}+{P}{\left({F}\right)}-{2}{P}{\left({E}{F}\right)}\)

Hence proved!

\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}\)

Now,

\(\displaystyle{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}-{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}\)

Since \(\displaystyle{P}{\left({E}{F}^{{c}}\cap{E}^{{c}}{F}\right)}={0}\)

\(\displaystyle\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}{F}^{{c}}\right)}+{P}{\left({E}^{{c}}{F}\right)}\)

Also, \(\displaystyle{P}{\left({E}{F}^{{c}}\right)}={P}{\left({E}\right)}-{P}{\left({E}{F}\right)}\) and \(\displaystyle{P}{\left({E}^{{c}}{F}\right)}={P}{\left({F}\right)}-{P}{\left({E}{F}\right)}\Rightarrow{P}{\left({E}{F}^{{c}}\cup{E}^{{c}}{F}\right)}={P}{\left({E}\right)}+{P}{\left({F}\right)}-{2}{P}{\left({E}{F}\right)}\)

Hence proved!